o****e
发帖数: 80 | 1 X(t)=W(t) and Y (t) = int_0_t{W(s)}ds.
if X(t)Y(t) = \int_0^t W(s)^2ds + \int_0^tY(t)dW(t)
E[X(t)Y(t)] =E[ \int_0^t W(s)^2ds + \int_0^tY(t)dW(t) ]
=\int_0^t E[W(s)^2] ds + \int_0^t E[Y(t)] dW(t) ???? 这一步对吗？根据什么
定理？好像不是 ito isometry把？ |
a*******h
发帖数: 123 | 2 Fubini?
【在 o****e 的大作中提到】
: X(t)=W(t) and Y (t) = int_0_t{W(s)}ds.
: if X(t)Y(t) = \int_0^t W(s)^2ds + \int_0^tY(t)dW(t)
: E[X(t)Y(t)] =E[ \int_0^t W(s)^2ds + \int_0^tY(t)dW(t) ]
: =\int_0^t E[W(s)^2] ds + \int_0^t E[Y(t)] dW(t) ???? 这一步对吗？根据什么
: 定理？好像不是 ito isometry把？
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t********t
发帖数: 1264 | 3 you can work this out by either Fubini theorem or Ito isometry.
The result is t^2/2 |
r*******y
发帖数: 290 | 4 how to work out using ito isometry?
【在 t********t 的大作中提到】
: you can work this out by either Fubini theorem or Ito isometry.
: The result is t^2/2
|
t********t
发帖数: 1264 | 5 E[ \int_0^t W(s)^2ds ]=E[(\int_0^t W(s)dW(s))^2] |
r*******y
发帖数: 290 | 6 nice, thanks
【在 t********t 的大作中提到】
: E[ \int_0^t W(s)^2ds ]=E[(\int_0^t W(s)dW(s))^2]
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b***k
发帖数: 2673 | 7 Then how to continue to get the answer?
I think you make the problem even harder if you work on it in this way.
【在 t********t 的大作中提到】
: E[ \int_0^t W(s)^2ds ]=E[(\int_0^t W(s)dW(s))^2]
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j*****4
发帖数: 292 | 8 yes,this method leads to a more complex track.
The following is what I did:
E(int_t^0 W_s dW_s)^2
=1/4 E(W_t^2-t)^2
=1/4 E(W_t^4-2tW_t^2+t^2)
recall that
E(W_t^2)=t
E(W_t^4)=3t^2( ...=E(int_t^0 d(W_s^4)) then use Ito lemma)
so the answer is t^2/2.
【在 b***k 的大作中提到】
: Then how to continue to get the answer?
: I think you make the problem even harder if you work on it in this way.
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b***k
发帖数: 2673 | 9 I got it the same way, that's why I said this approach may not be a good one.
【在 j*****4 的大作中提到】
: yes,this method leads to a more complex track.
: The following is what I did:
: E(int_t^0 W_s dW_s)^2
: =1/4 E(W_t^2-t)^2
: =1/4 E(W_t^4-2tW_t^2+t^2)
: recall that
: E(W_t^2)=t
: E(W_t^4)=3t^2( ...=E(int_t^0 d(W_s^4)) then use Ito lemma)
: so the answer is t^2/2.
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